Question

Draw a circle of radius 3 cm. Take two points P and Q on one of
its extended diameter each at a distance of 7 cm from its
Centre. Draw tangents to the circle from these two points.
​

Answer 1

${\bigstar}$SOLUTION${\bigstar}$

CONSTRUCTION:

➤Draw a circle of radius 3cm .

➤Draw a line segment of length 7cm from the center of the circle O

➤With OP as line segment ,draw a perpendicular bisectors to meet at the point "M" which is perpendicular to line segment .

➤With 'M' as center ,draw a circle to meet at the points 'A' and 'B' at the another circle of radius 3cm

➤Join A and P ,B and P .

➤Thus ,we obtained the two tangents .

${\bigstar}$TO FIND${\bigstar}$

• Length of the tangents

by Pythagoras theorem,

$\boxed{\sf{Square of hypotenuse = Square of base + square of perpendicular side }}$

➠$\boxed{\sf{ OP² = OA² + AP² }}$

➠7² = 3² + AP²

➠AP² = 7² - 3²

➠AP² = 49 - 9

➠AP² = 40

➠AP = √40

➠AP = 6.3 (app.)

Therefore ,

the length of tangents is

$\boxed{\sf{ AP = 6.3 cm }}$

Answer 2

$\huge\underline\mathfrak\pink{Answer⤵}$

Solution,

CONSTRUCTION:

➤Draw a circle of radius 3cm .

➤Draw a line segment of length 7cm from the center of the circle O

➤With OP as line segment ,draw a perpendicular bisectors to meet at the point "M" which is perpendicular to line segment .

➤With 'M' as center ,draw a circle to meet at the points 'A' and 'B' at the another circle of radius 3cm

➤Join A and P ,B and P .

➤Thus ,we obtained the two tangents .

To find,

Length of the tangents

by Pythagoras theorem,

$\boxed{\tt{Square of hypotenuse = Square of base + square of perpendicular side }}$

➠$\boxed{\tt{ OP² = OA² + AP² }}$

➠7² = 3² + AP²

➠AP² = 7² - 3²

➠AP² = 49 - 9

➠AP² = 40

➠AP = √40

➠AP = 6.3 (app.)

Therefore ,

the length of tangents is

$\boxed{\tt{ AP = 6.3 cm }}$